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Chapter 3 Problems
3.1 After the population reaches equilibrium, the chance of any haplotype would be equal to any other (ie. AB/AB=.25 Ab/aB=.25 ab/ab=.25 aB/Ab=.25). 3.2) two markers 1cM apart: a) the probability that there will be remobination between these two markers in one generation is 1% (= 0,01). One centimorgan corresponds to 1% of recombination between sister chromatids of two observed locus locations. (at 0cM no recombination between locuses occurs) b) the probability that there won't be a genetic recombination between these two locuses in one generation is 99% (=0,99). c) the probabily of recombinatin between these two locuses in any generation is 0,01. The probability that recombination won't happen is 1-probabilíty that recombination will happen. In this case 0,99 in a generation. This number can be multiplied by the number of generations. d) the probabily of recombination in *10 generations is 0,1 *20 generations is 0,2 *30 generatins is 0,3 *40 generations is 0,4 *50 generation is 0,5 the probability that recombination won't occur is than: 1-0,1 =0,9 for ten generations, 1-0,2= 0,8 for twenty generations, 1-0,3= 0,7 for thirty generations, 1-0,4= 0,6 for fourty generations and 0,5 for fifty generations. 3.2) A distance of 1cM translates to a probability of 1% of a recombination event. a) So, for the first generation there is a 1% chance. b) If there's a 1% chance of recombination, there is a 99% chance of no recombination. c) X = distance in cM, n = generation number Probability (% chance) of no recombination = (100 - X) - (n - 1) d) for n = 10, we have (100 - 1) - (10 - 1) = 90, or 90%. for n = 20, we have (100 - 1) - (20 - 1) = 80% , for n = 30 we would have a 70% chance, for n = 40 we have a 60% chance, and for n = 50 we would have a 50% chance. 3.6. sequencing gel: The sequence of the fragment shown on the radiograph sequencing gel (figure 3.25) is: CCTTATCTCTTGATAGGGAACCGGTTTCCCGGC. There is a complementary region (underlined in the sequence), that might possible create hairpin loops. This region will probably encode a functional for of an RNA molecule. 3.7. SOLID Results a) Sequence of the parent strand fragment is 5'-TACGACGACTGACGACTACTCTTCCGATCCACCAGTC-3' b) The difference in the SOLID map would differ in read positions 14 & 15 for primer rounds 2, 3 & 4. This causes read position 14 to glow red (TA) instead of green (TG) and read position 15 to glow blue (AA) instead of yellow (GA). That is the only location that they would differ. 3.8 Cost a.) The rate of change of the cost of sequencing over years 2005-2007. In January 2005, a megabase of DNA was about 974.16 dollars and in January 2007 a megabase of DNA was about 522.71 dollars. Assuming cost decreases linearly, the rate of change is -451.45 dollars/2 years or -225.73 dollars/year. b.) From 2008-2010, the rate of cost change was (from Jan to Jan) $ 102.13 to $0.52. This calculates to -50.81 dollars per year. c.) The cost of determining a human sized genome at 10x coverage would be about be greater than $46,774 . This is because 1 genome would have cost $46,774. Also when you are talking about coverage you will have multiple fragments covering a region of the DNA (to determine the sequence) and if you have multiple fragments you should also have multiple genomes that have been fragmented (so that fragments overlap). Since coverage is C = NL/G. A coverage of 10 would mean that NL are 10 times greater than the genome length. I would guess that a genome with 10x coverage would cost $460,774.